Mathematical induction on a finite subset of $\mathbb{N}$

Question

I'm trying to prove the following using induction (or well-ordering principle). (Induction says that whenever $ 0 \in S $ and $ n \in S \implies (n+1) \in S$, then $S = \mathbb{N} $. )

Let $ \{ P_{k} \}$ be a sequence of propositions such that
$P_{1}$ is true
The truth of $P_{i-1} $ implies the truth of $P_{i}$ for all $(i-1) \in \{ 1,2,...(n-1) \} $

Then show that ${P_{k}}$ is true for all $ k \in \{ 1,2,...n \} $

Clearly, this is a modified form of induction (perhaps a weaker form) restricted to the first $n$ naturals, so the knowledge of induction should easily allow me to prove this. Yet I am stuck, what's the trick?

Answer 1

On easy way to use the induction principle is to create a new proposition which hold for all natural numbers.

Consider $Q_k$ to be the proposition $P_k$ if $k\leq n$, and other wise i.e. if $k>n$ then let $Q_k$ be the statement $k>n$. Now you may prove that $Q_k$ hold using induction. Notice that you will need to divide into cases in the induction step depending on if the number you do incution over is greater than $n $ or not.

Answer 2

Now we have the truth of $P_{i-1}$ implies the truth of $P_i$ for all $(i-1) \in \{ 1,2,...,n \}$, and we repeat.

Answer 3

We use the Well-Ordering Principle.

Let $T = \{k \in \{1, 2, 3, \ldots, n\} \mid P_k~\text{is false}\}$. We will show that $T = \emptyset$.

Suppose otherwise. Then since $T \subseteq \mathbb{N}$ and $T \neq \emptyset$, there exists a least element $m \in T$ such that $P_m$ is false. Since $P_1$ holds, $m \neq 1$. Since $m > 1$ and $m - 1 < m \leq n$, $P_{m - 1}$ holds. However, $P_{m - 1} \implies P_m$ holds. Thus, $m \notin T$, contrary to our assumption that $m$ is the least element of $T$. Hence, $T$ does not have a least element, so $T = \emptyset$. Therefore, if $P_1$ holds and $P_{k - 1} \implies P_k$ for all $k \in \{1, 2, 3, \ldots, n - 1\}$, $P_k$ holds for each $k \in \{1, 2, 3, \ldots, n\}$.